The roots of the equation […] = 0 are

The correct answer is Option (3) → \(2, 3, -5\)

The determinant is a cubic in x, so it has three roots. Values of x that make two rows identical give roots at once.

At x = 2, the first two rows are both (2, 2, 3), so the determinant is 0.

At x = 3, the last two rows are both (2, 3, 3), so the determinant is 0.

For the third root, the operation \(C_1 \to C_1 + C_2 + C_3\) makes every row start with \(x + 5\):

    \(\displaystyle \Delta = \begin{vmatrix}x + 5 & 2 & 3\\ x + 5 & x & 3\\ x + 5 & 3 & x\end{vmatrix}\)

    \(\displaystyle \Delta = (x + 5)\begin{vmatrix}1 & 2 & 3\\ 1 & x & 3\\ 1 & 3 & x\end{vmatrix}\)

So x = −5 is also a root, and

    \(\Delta = (x + 5)(x - 2)(x - 3)\)

Check at x = 0: \(\begin{vmatrix}0 & 2 & 3\\ 2 & 0 & 3\\ 2 & 3 & 0\end{vmatrix} = 0 - 2(0 - 6) + 3(6 - 0) = 30\), and \((5)(-2)(-3) = 30\).